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The solution is $x=50^{\circ}$.

How to prove $x=50^{\circ}$ without trigonometry?

VividD
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kong
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2 Answers2

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Find $K$ on $\overline{BD}$ such that $\angle KAB = 10^\circ$. Then $\angle AKB = 160^\circ = 2\cdot\angle ACB$; this makes $K$ the center of the circumcircle of $\triangle ABC$. Thus, $\overline{KB}\cong\overline{KC}$, whereupon $\angle KCB = 20^\circ$. Note also that $\triangle ACK$ must be equilateral. (Why?)

We also have that $\angle DKA = 20^\circ = \angle DAK$. This makes $\triangle DAK$ isosceles, so that $\overline{CD}$ bisects $\angle ACK = 60^\circ$. The result follows. $\square$

Blue
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    (+1) Think I got that. Have you used the identified $k$ to be the centre of the circle because of the arrow theorem? – Karl Feb 21 '15 at 11:22
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    @Karl: Yes. Because the angle at $K$ looking at $\overline{AB}$ is twice the angle at $C$ looking at $\overline{AB}$, the Arrow Theorem implies that $K$ is the circumcenter of $\triangle ABC$. – Blue Feb 21 '15 at 11:34
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    I have learnt something in attempting to answer this and from seeing your answer. It deserves to be accepted as the solution. Thanks. – Karl Feb 21 '15 at 11:37
  • @abiessu: Isn't $\angle CAD = 40^\circ$ in the question's diagram? – Blue Feb 22 '15 at 05:28
  • @Blue: of course, if I miss a simple flip and slight rotation, I would miss the correctness of the answer... – abiessu Feb 22 '15 at 20:00
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This is incorrect and just left for reference

The strategy to take appears to be to find all the angles in the centre at $D$ in terms of $x$ and then equate to $360$ but I don't get $x=50$

$\angle BDC=180-20-x=160-x$

$\angle ADB=180-30-10=140$

Use the largest triangle to find $\angle ACD$

$\angle ACD =180-70-30-x=80-x$

This means $\angle ADC =180-40-(80-x)=60+x$

At the centre you have $160-x+140+60+x=360$

The $x$ cancel and I briefly checked that $x=60$ is a solution as well.

See comments below as to why this fails

Karl
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    If $x$ cancels, then you can't solve for it. Moreover, it certainly isn't possible to find an alternate solution: $\triangle ABC$ and $\triangle ADC$ are uniquely determined by the angles at $A$ and $B$ (and the length of $\overline{AB}$); consequently, the entire figure is fixed, so that $x$ has exactly one value. (A GeoGebra sketch seems to confirm that $x=50^\circ$, but I don't have a non-trigonometric proof.) – Blue Feb 21 '15 at 09:23
  • I had horrible feelings about this one. Shall I leave my misendeavour or delete it? – Karl Feb 21 '15 at 09:29
  • Put comment on showing my error. Thanks. – Karl Feb 21 '15 at 09:37
  • You're a teacher, so you could transform this to a teachable moment. Edit to declare the answer incorrect, and then explain why it is incorrect. After all, many students are bound to make this kind of mistake themselves. (Ah ... You've edited just as I've written this. I guess I'm more persuasive than I realized. ;) – Blue Feb 21 '15 at 09:37
  • It is the uniquely determined bit that tripped me up. I need to think that through. – Karl Feb 21 '15 at 09:40